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8 points) Find a basis for subspace H = Span { b1,bz,b3,b4,bs}, where2 7 b1 = -4 bz 55 05 3 3 4 313 2 13 bs ~7 915 -26 -22 17 -27-3 ~9 1 8b3b4...

Question

8 points) Find a basis for subspace H = Span { b1,bz,b3,b4,bs}, where2 7 b1 = -4 bz 55 05 3 3 4 313 2 13 bs ~7 915 -26 -22 17 -27-3 ~9 1 8b3b4

8 points) Find a basis for subspace H = Span { b1,bz,b3,b4,bs}, where 2 7 b1 = -4 bz 55 0 5 3 3 4 3 13 2 13 bs ~7 9 15 -26 -22 17 -27 -3 ~9 1 8 b3 b4



Answers

(a) find $n$ such that rowspace $(A)$ is a subspace of $\mathbb{R}^{n}$, and determine a basis for rowspace $(A) ;$ (b) find $m$ such that colspace $(A)$ is a subspace of $\mathbb{R}^{m},$ and determine a basis for colspace $(A)$. $$A=\left[\begin{array}{lll} 1 & 2 & 3 \\ 5 & 6 & 7 \\ 9 & 10 & 11 \end{array}\right]$$

In this video, we will be finding a basis for the road space and a basis for the column space of the following matrix, which all reveal here Now I've called this Matrix P and let's start out with finding a basis for the road space. Now remember, for finding a basis for the road space, we can use the following to fax to kind of streamline our our efforts. So the first fact this that if 22 matrices are row equivalent So let's AP ISRO equivalent to some matrix p prime and that implies if they're row spaces are the same. So the road space of P is equal to the road space of Pete crime. The second fact that we can use it was kind of going to streamline our methods to finding a basis for the roast base of this guy Is that the non zero rows of a row echelon form matrix of the non zero rose of a matrix in a row Echelon form the Nancy roads of a major turn Rochel on form are linearly independent are linearly independent, so weaken rove reduce this guy to some row echelon form of the Matrix p and take the non zeros And those non zeros will form a basis for the road space of P. So let's vote Reduce this guy. So take a second and positive video and give a give a try it Go give give ago it road reducing this matrix people. So I'm assuming you've given it a try. So I'm going to reveal the Siri's of RO equivalent matrices of P that lead us to the to a row echelon form of P right now. So after a series of elementary row operations, we ro reduce the matrix to these matrices and eventually ended up at the following row Echelon form or a row echelon form of the matrix P. Go ahead and check this and make sure that you understand what we did here. So I'm assuming you've understood what we've done here. So now we know that the row space of P prime this matrix that is a row echelon form of the matrix P is the same as the road space of P. Now the road space of P prime is the span of the row vectors of these row factors. So it's the span. It's the span of these three row vectors are just going to call. These row vectors are one are to R three and R four. So it'll have to write these long row vectors here. So it's a span of our one hard to are three and our four now r three and r four R zero vectors And remember, we can always remove or it's OK if you don't remember because I this is sometimes this is sometimes Ah, nice. A little trick to use and totally okay if you don't remember. But we are always a nice little trick that we can kind of always having are back in our back pocket is we can always remove zero vectors from a spanning set and we get a new set of vectors. Once we remove those, your vectors who span is the same as the span off vectors. With these your vectors not removed. So because these your vectors are just linear combinations of the other two vectors in these and the set namely zero times are one plus zero times are two plus zero times R three and r three, of course, is also a zero vector. So it zero times are one plus zero times out to put zero times are four. So R three and R four are linear combinations of the other vectors in this spanning set. So we get a new set of vectors whose span we get a new set of vectors, namely the two vectors R one and R two, whose span is the same as the original set of vectors, with zero vectors not removed. So since this span is equal to the road space of P, we also know that this ban is equal to the road space of P. And moreover, moreover, remember, the non zero vectors in a row echelon form matrix are linearly independent. So we know that r one and r two the vectors zero excuse me One comma, two common negative one comma three and zero comma, zero comma zero comma one. These two vectors are linearly independent and they're span is equal to the road space of P. So we found of basis for the road space of P. And that's really nice because we all we had to do was road of sin matrix toe a row echelon form, and then take the non zero bows and we got our basis or a basis for the aerospace of P. No, the road space of P is of course, a span of its row vectors. The span of its row vectors and the span of its row vectors are any linear combination of these row vectors. In any linear combination of these row vectors, we'll have four rial number components. So therefore, we can say that the road space of P is a subspace of our four, the set of all vectors containing four real number components. So if you want to find the column space of P, we could listen a variety of ways. If there's more than one way to find the common space in the road space, because our in goal is to find a linear linearly independent set s a linearly Independence, that s who span who span is equal to the Collins face of P and again is multiple ways to do this. But one nice way since we already have a row echelon form of P. One Nice way is to locate the pivot cons of p. So locate Piva Cons, API locate pivot columns, calls of P and then form a basis, a basis form a basis for the confidence of P. By taking the pitta cons of p. Pity it calms of P. That is to say that the pivot columns of P form a basis for the column space of P, and the reason for this is a little as a little complex in the explanation and goes a little bit beyond the scope of this video. It's not that you couldn't it's not that we couldn't understand it. It's just that it would take a little bit too long to explain why this is true. But the argument the core of the argument has to do with the falling. If two matrices of Roe equivalent say a and B a row equivalent, then the solution to the equation a X is equal to 00 vector and be Times X is equal to the zero factor. The X values that make these two equations true are the same. Are the same for two row equivalent matrices weaken. Rephrase this as the to Roe equivalent matrices have the same linear dependency relationships. No, again, I have ah, little lecture video. I will have a little extra video in my office, our section of my profile, if you're interested mawr and into the the details of why, why This allows us to take the pivot Collins of P and former basis for the common space of P using those pivot call. But anyways so when we reduced gloomy Rover DUSA Matrix to a row echelon form of a matrix. So when we wrote reduced P to a Rolex swan, former peace AP Prime. This allowed us to locate the pivot columns of P in the pivot columns of P correspond exactly to the pivot calms API Prime and the people calling for P prime are the columns with leading entries. This column right here and this column right here. This is a leading entry, and this is a leading entry or sometimes called a leading one or also called a pivot position. So all we have to do is take this column right here. The column that corresponds with the Pivot column and pee prime because the pivot comes and pee primer. Exactly. The Pivot columns and P and this cholera here the column that corresponds with the column that contains a leading one in P Prime, the column in P the Collins and P the correspond exactly with the columns in P prime that contain leading once. But we have to be careful. Warning. We have to take the Collins in pea to form our basis. We do not want to take the columns in p crime, the former basis. We want to take Occam's in pea to form our basis. The column vectors. Rather so we have. This tells us, by this method. By this method, we have that the two basis vectors for the column space of P or rather, the two and only two basis vectors for the con space api are the vectors. One comma, three comma, one comma, five and three comma, five comma negative one comma seven. And these two vectors form a basis for the common space of P. And, moreover, again, since the base are since the college base of P is the span of all the contractors API, in the span of all the contractors appear all linear combinations of these vectors. With four real number components, we can say that the column space of P, we can say that the column space of PD is a subspace of our four because each vector in the cost base of Fee P has four real number components, and so therefore, it must be a subspace of our for the set of all vectors containing four of your number components.

Yeah. Okay. Suppose. Were given three vectors 1, -4 negative three, And then 20 to -2 in two negative 132 Okay. And given these three vectors, uh I want to find a basis. Okay, for the subspace that is spanned by them. Okay, subspace of R. For So we consider a subspace of R4. That is spanned by these three factors. And I want to find a basis for that subspace. Well, recall what is the basis? Well, it requires that the basis vectors are linearly independent. Okay. And that they span the space. Okay. But we already given that they span the subspace. Okay. That that's the premise of the question. Okay, So we don't have to worry about the spanning part of the basis requirements. We all we have to do is we need to make sure that, you know, these three vectors are independent. If these three vectors are linearly independent. Okay. Then we're done. Right, Because by definition they span the space that they span. So as long as these vectors are linearly independent, then they form a basis. Okay, So how do we check whether these vectors are linearly independent? Well, the easiest way over here is to simply see that um you know, like there are no ways to there are no nontrivial linear combinations that you zero. Okay. So for example, over here, right? You can let's look at vector to vector three. Okay, so you can see that if I were to subtract, the only way for for us to get to zero in the first entry on the top here is to subtract them from each other, right? Like like uh I get zero at the top but then none of the other entries would be zero, Right? Because if you subtract them, even though the first entry cancels out, none of the other entries who cancel out, so therefore there does not exist. Okay, any nontrivial linear combination of this? Uh these two vectors that would use zero, so therefore these two vectors are linearly independent. Okay, now let's look at the first two vectors. Okay? Um if I the only way okay for us to scale uh for us to get a zero in the first entry is to, you know, like multiply the first one by a factor of two. And then a subtract. Okay, so, well, only up to a scalar constant, but that doesn't really matter. Um So over here I have taught I have a zero at the top. Okay. But then over here you would see that even though I get zero from the top by multiplying this by two and then subtracting uh none of the other entries, Okay, would be zero. Okay, So therefore there are no uh nontrivial linear combinations of these two vectors that would use zero, either and last, but not least. Let's look at the one and 3. Let's look at vectors more than three. And again, you see that if I scaled the first factor by two, and I and I and I subtract the third factor from it, that would get the first entry to zero, but none of the other entries would be zero. So therefore, we conclude that uh none. Uh you can you can basically you get the idea of this exercise right? Which is to just try to figure out is to try to to uh figure out a justification for the fact that there are no nontrivial linear combinations of these three factors that would yield the zero vector, and therefore they are linearly independent, and therefore they span the space that they span and they form the basis. So the answer okay. For the basis, the simplest answer that you can find okay without doing all that that, you know, like reduction, you know, like us or like those matrix manipulations things, is to just say that aha. The basis is the fault is given to us already. The basis is just 11 negative four, negative 3 to 0 to negative two and two negative 132 Okay.

Okay. In this question, we will find the basis of the subspace span by these vectors. So this is a quiver tow asking what is a column space off this matrix when you just squish all of these colored pictures together? So this column back the girls yet some guys here goes here, goes here, goes here. Does it really matter what water it is? But but I'll just take a funeral. So then the first thing that we have to do is to find to convert this into right echelon for so the first couple off steps that will do is take a road to you, add one that she wrote to will take Row three and add two tons of road one. In that interview, I three three and then we're full. You was tracked. Why? It has wrote one. Bring that into road for so performing the steps 120 minus 131 plus one gives you zero to my street. Get it minus 10 clothes to give to four months. One gives you three, three months. Eight gives you negative five now very three plus words. You so negative Two plus two gives you zero four minus one gives you 30 My sixties Negative six negative seven miles to give me make you mine and nine plus six kids. A kitty. Okay, now rough Full minus 51 So five minus 50 six minus 10. Giving negative pole eight minus zero gives you eight. Seven plus five. Getting too. No. Sorry. Well, and by minus 15 gives you maybe 10. So now, right oppressed That perform is row three and add three. Title wrote to going down into row three and wrote full will subtract four tunnel road to connect to work for So we get 1 to 0 minus 130 Advice 123 Negative five. And you're so right. Three plus road to sew. Three plus three miles st zero to negative six plus six is your negative mind plus nine zero 15 minus 15. Now I wrote for my forward to Is there a negative? Four plus four Q zero eight minus eight is you 0 12 miles. Talk gives you zero, but this is a regular for my sake. Zero on dhe negative. 10 plus 20 gives you 10. So you see, we only have three columns, right? Yeah. So how How homes waste is the space to span by These three letters stood in the basis for this subspace. Ah, one minus one minus 25 23 months. 16 and three. Negative, 89 and five.

Okay, you this question? We want to show that the settle Victor's B one b two b three are in fact at basis for hatch on that, actually supporting. So, what we do in this situation is we simply creating matrix where we write metrics A has bean B one, B two, B three and finally X. All we need to do is reduce this into road. Just wrote a song form on DDE. In theory, if it is in fact a basis if you want me to be three are a basis. Then we should have freed. Give the columns. So here I create apartment script Where this is the matrix, This is Matrix. Is this so the first way we'll have minus 68 minus time. Four and second road will have full ministry five and seven and so on. So on. And so this step performs of registration form, and here we can see that the reduced for extreme form does, in fact, have 123 minutes now to find the xB coordinates we need to look at is simply this last call 352 And in fact, xB is, in fact, three. Why don't you


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